Simplify and expand the following expression: $ \dfrac{1}{5n + 45}- \dfrac{1}{4n - 8}+ \dfrac{1}{n^2 + 7n - 18} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{1}{5n + 45} = \dfrac{1}{5(n + 9)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4n - 8} = \dfrac{1}{4(n - 2)}$ We can factor the quadratic in the third term: $ \dfrac{1}{n^2 + 7n - 18} = \dfrac{1}{(n + 9)(n - 2)}$ Now we have: $ \dfrac{1}{5(n + 9)}- \dfrac{1}{4(n - 2)}+ \dfrac{1}{(n + 9)(n - 2)} $ The least common multiple of the denominators is: $ 20(n + 9)(n - 2)$ In order to get the first term over $20(n + 9)(n - 2)$ , multiply by $\dfrac{4(n - 2)}{4(n - 2)}$ $ \dfrac{1}{5(n + 9)} \times \dfrac{4(n - 2)}{4(n - 2)} = \dfrac{4(n - 2)}{20(n + 9)(n - 2)} $ In order to get the second term over $20(n + 9)(n - 2)$ , multiply by $\dfrac{5(n + 9)}{5(n + 9)}$ $ \dfrac{1}{4(n - 2)} \times \dfrac{5(n + 9)}{5(n + 9)} = \dfrac{5(n + 9)}{20(n + 9)(n - 2)} $ In order to get the third term over $20(n + 9)(n - 2)$ , multiply by $\dfrac{20}{20}$ $ \dfrac{1}{(n + 9)(n - 2)} \times \dfrac{20}{20} = \dfrac{20}{20(n + 9)(n - 2)} $ Now we have: $ \dfrac{4(n - 2)}{20(n + 9)(n - 2)} - \dfrac{5(n + 9)}{20(n + 9)(n - 2)} + \dfrac{20}{20(n + 9)(n - 2)} $ $ = \dfrac{ 4(n - 2) - 5(n + 9) + 20} {20(n + 9)(n - 2)} $ Expand: $ = \dfrac{4n - 8 - 5n - 45 + 20}{20n^2 + 140n - 360} $ $ = \dfrac{-n - 33}{20n^2 + 140n - 360}$